Answer:
basketball player scored 40 points in a game. The number of three-point field goals the player made was 22 less than three times the number of free throws (each worth 1 point). Twice the number of two point field goals the player made was 11 more than the number of three point field goals made. Find the number of free-throws, two point field goals, and three point field goals that the player made in the game.
Once I have the equations I understand how to do the problem, but I am obviously doing something wrong because each time I write out the equations I end up getting fractions.
Answer:
4x - 3
Step-by-step explanation:
(7x - 2) - (3x - 5)
Open the bracket and observe the signs,
7x -2 - 3x + 5 ( when the minus outside the bracket multiplies minus in the bracket, it becomes plus)
Collect like terms
7x - 3x - 2 + 5
4x - 3
Answer:
we conclude that:
Step-by-step explanation:
Given the expression
cancel the common factor: 2
cancel the common factor: p
cancel the common factor: 2p+1
Expanding
Thus, we conclude that:
Answer:
Since both terms are perfect squares, factor using the difference of squares formula,
a
^2
−
b
^2
=
(
a
+
b
)
(
a
−
b
)
where
a
=
y
and
b
=
6
(
y
+
6
)
(
y
−
6
)
The sector (shaded segment + triangle) makes up 1/3 of the circle (which is evident from the fact that the labeled arc measures 120° and a full circle measures 360°). The circle has radius 96 cm, so its total area is π (96 cm)² = 9216π cm². The area of the sector is then 1/3 • 9216π cm² = 3072π cm².
The triangle is isosceles since two of its legs coincide with the radius of the circle, and the angle between these sides measures 120°, same as the arc it subtends. If b is the length of the third side in the triangle, then by the law of cosines
b² = 2 • (96 cm)² - 2 (96 cm)² cos(120°) ⇒ b = 96√3 cm
Call b the base of this triangle.
The vertex angle is 120°, so the other two angles have measure θ such that
120° + 2θ = 180°
since the interior angles of any triangle sum to 180°. Solve for θ :
2θ = 60°
θ = 30°
Draw an altitude for the triangle that connects the vertex to the base. This cuts the triangle into two smaller right triangles. Let h be the height of all these triangles. Using some trig, we find
tan(30°) = h / (b/2) ⇒ h = 48 cm
Then the area of the triangle is
1/2 bh = 1/2 • (96√3 cm) • (48 cm) = 2304√3 cm²
and the area of the shaded segment is the difference between the area of the sector and the area of the triangle:
3072π cm² - 2304√3 cm² ≈ 5660.3 cm²
